The coefficient self inductance of a solenoid is 0.18 mH. If a core of soft iron of relative permeability 900 is inserted, then the coefficient of self-inductance will becomes

Question

The coefficient self inductance of a solenoid is 0.18 mH. If a core of soft iron of relative permeability 900 is inserted, then the coefficient of self-inductance will becomes (A) 0.0002 mH (B) 0.006 mH (C) 5.4 mH (D) 162 mH

Tardigrade Admin · Accepted Answer

We know that, L=μ ni ∴ (L2/L1)=(μ /μ 0) (as n and i are same) ⇒ L2=(μ /μ 0)× L1=μ rL1 =900× 0.18=162 text mH

Q. The coefficient self inductance of a solenoid is 0.18 mH. If a core of soft iron of relative permeability 900 is inserted, then the coefficient of self-inductance will becomes

0.0002 mH

0.006 mH

5.4 mH

162 mH

We know that, $L = μ ni$ $∴$ $\frac{L _{2}}{L _{1}} = \frac{μ}{μ _{0}}$ (as n and i are same) $\Rightarrow$ $L_{2} = \frac{μ}{μ _{0}} \times L_{1} = μ_{r} L_{1}$ $= 900 \times 0.18 = 162 m H$

Q. The coefficient self inductance of a solenoid is 0.18 mH. If a core of soft iron of relative permeability 900 is inserted, then the coefficient of self-inductance will becomes

0.0002 mH

0.006 mH

5.4 mH

162 mH

We know that, L=μni ∴ L1​L2​​=μ0​μ​ (as n and i are same) ⇒ L2​=μ0​μ​×L1​=μr​L1​ =900×0.18=162mH

We know that, $L = μ ni$ $∴$ $\frac{L _{2}}{L _{1}} = \frac{μ}{μ _{0}}$ (as n and i are same) $\Rightarrow$ $L_{2} = \frac{μ}{μ _{0}} \times L_{1} = μ_{r} L_{1}$ $= 900 \times 0.18 = 162 m H$